Answer:

Explanation:
<u>Given Data:</u>
Acceleration due to gravity = g = -9.8 m/s²
Maximum Height = h = 0.490 m
At h,
= 0
<u>Required:</u>

<u>Formula:</u>

<u>Solution:</u>
Put the givens
![2 (-9.8) (0.490) = (0)\² - v_o^2\\\\-9.604=-v_o^2\\\\9.604=v_o^2\\\\Take \ sqrt\ on \ both \ sides\\\\\sqrt{9.604}=v_o^2\\\\3.1 \ m/s=v_o\\\\v_o=3.1\ m/s\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=2%20%28-9.8%29%20%280.490%29%20%3D%20%280%29%5C%C2%B2%20-%20v_o%5E2%5C%5C%5C%5C-9.604%3D-v_o%5E2%5C%5C%5C%5C9.604%3Dv_o%5E2%5C%5C%5C%5CTake%20%5C%20sqrt%5C%20on%20%5C%20both%20%5C%20sides%5C%5C%5C%5C%5Csqrt%7B9.604%7D%3Dv_o%5E2%5C%5C%5C%5C3.1%20%5C%20m%2Fs%3Dv_o%5C%5C%5C%5Cv_o%3D3.1%5C%20m%2Fs%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Part a
Answer: 17.58 km/h

Total Distance =10 km
Total time =0.5689 h

Part b
Answer: 17.626 km/h

Total Distance =42.195 km
Total time =2.3939 h

Yes I can do you want me to
An object is thrown horizontally from the open
window of a building. If the initial speed of the
object is 20 m/s and it hits the ground 2.0 s later,
from what height was it thrown? (Neglect air
resistance and assume the ground is level.)
Initial vertical velocity = 0
formula: s = ut + (1/2)*g*t^2
s = 0 + 1/2*9.8*2^2
s = 9.8*2 = 19.6 m
Answer:
I don't know I'm sorry I will tell you another answer asks me